Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int gauss(vector<int>& v, vector<pair<int, int> >& ans, vector<int>& mb) {
  int cur = 0;
  for (int b = 30; b >= 0; --b)
    if (cur < v.size()) {
      for (int i = cur; i < v.size(); ++i)
        if (v[i] & (1 << b)) {
          if (i != cur) {
            swap(v[i], v[cur]);
            ans.push_back(pair<int, int>(i, cur));
            ans.push_back(pair<int, int>(cur, i));
            ans.push_back(pair<int, int>(i, cur));
          }
          break;
        }
      if (v[cur] & (1 << b)) {
        for (int i = 0; i < v.size(); ++i)
          if (i != cur && (v[i] & (1 << b))) {
            v[i] ^= v[cur];
            ans.push_back(pair<int, int>(i, cur));
          }
        mb.push_back(1 << b);
        ++cur;
      }
    }
  return cur;
}
int main() {
  scanf("%d", &n);
  vector<int> x(n), y(n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &y[i]);
  vector<pair<int, int> > ans;
  ans.reserve(1000000);
  vector<int> mb;
  int b = gauss(x, ans, mb);
  vector<int> y0 = y;
  for (int i = 0; i < y.size(); ++i) {
    for (int j = 0; j < b; ++j)
      if ((y[i] & x[j]) >= mb[j]) {
        y[i] ^= x[j];
        if (i >= b) ans.push_back(pair<int, int>(i, j));
      }
    if (y[i]) {
      printf("-1\n");
      return 0;
    }
  }
  y0.resize(b);
  vector<pair<int, int> > ansy;
  vector<int> mby;
  gauss(y0, ansy, mby);
  y = y0;
  for (int i = 0; i < y.size(); ++i) {
    if ((x[i] & y[i]) >= mb[i]) {
      y[i] ^= x[i];
    } else {
      ans.push_back(pair<int, int>(i, i));
    }
    for (int j = i + 1; j < b; ++j)
      if ((y[i] & x[j]) >= mb[j]) {
        y[i] ^= x[j];
        ans.push_back(pair<int, int>(i, j));
      }
  }
  for (int i = (int)ansy.size() - 1; i >= 0; --i) {
    ans.push_back(pair<int, int>(ansy[i].first, ansy[i].second));
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); ++i)
    printf("%d %d\n", ans[i].first + 1, ans[i].second + 1);
  return 0;
}